For how many values of 'x' in the closed interval `[-4,-1]` is the matrix `[(3,-1+x,2),(3,-1,x+2),(x+3,-1,2)]` singular ? (A) `2` (B) `0` (C) `3` (D) `1`

Options

A. 0
B. 1
C. 2
D. 3

Answer

To determine how many values of 'x' in the closed interval `[-4, -1]` make the matrix

\[ A = \begin{pmatrix} 3 & -1 + x & 2 \\ 3 & -1 & x + 2 \\ x + 3 & -1 & 2 \end{pmatrix} \]

singular, we need to find when the determinant of the matrix is equal to zero.

### Step 1: Calculate the determinant of the matrix A

The determinant of a 3x3 matrix

\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

is given by:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix A, we have:

- \( a = 3 \)
- \( b = -1 + x \)
- \( c = 2 \)
- \( d = 3 \)
- \( e = -1 \)
- \( f = x + 2 \)
- \( g = x + 3 \)
- \( h = -1 \)
- \( i = 2 \)

Substituting these values into the determinant formula gives:

\[ \text{det}(A) = 3((-1)(2) - (-1)(x + 2)) - (-1 + x)(3(2) - (x + 2)(x + 3)) + 2(3(-1) - (-1)(x + 3)) \]

### Step 2: Simplify the determinant expression

Calculating each part:

1. \( (-1)(2) - (-1)(x + 2) = -2 + (x + 2) = x \)
2. \( 3(2) - (x + 2)(x + 3) = 6 - (x^2 + 5x + 6) = -x^2 - 5x \)
3. \( 3(-1) - (-1)(x + 3) = -3 + (x + 3) = x \)

Now substituting back into the determinant:

\[ \text{det}(A) = 3(x) - (-1 + x)(-x^2 - 5x) + 2(x) \]

Expanding this gives:

\[ = 3x + (1 - x)(x^2 + 5x) + 2x \]

### Step 3: Set the determinant equal to zero

Combine like terms:

\[ = 3x + 2x + (x^2 + 5x - x^3 - 5x^2) = 0 \]

This simplifies to:

\[ -x^3 - 4x^2 + 5x = 0 \]

Factoring out \( x \):

\[ x(-x^2 - 4x + 5) = 0 \]

### Step 4: Solve for x

Setting \( x = 0 \) gives one solution. Now we solve the quadratic:

\[ -x^2 - 4x + 5 = 0 \]

Multiplying through by -1:

\[ x^2 + 4x - 5 = 0 \]

Using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 20}}{2} = \frac{-4 \pm \sqrt{36}}{2} = \frac{-4 \pm 6}{2} \]

This gives:

1. \( x = \frac{2}{2} = 1 \)
2. \( x = \frac{-10}{2} = -5 \)

### Step 5: Identify valid solutions in the interval [-4, -1]

The solutions we found are:

- \( x = 0 \) (not in the interval)
- \( x = 1 \) (not in the interval)
- \( x = -5 \) (not in the interval)

The only valid solution in the interval `[-4, -1]` is \( x = -4 \).

### Conclusion

Thus, there is only **1 value** of \( x \) in the interval `[-4, -1]` that makes the matrix singular.

The answer is **(D) 1**.

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